Calcolatore di decomposizione in fratti semplici

Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{2} - 64}$$$

Factor the denominator: $$$\frac{1}{x^{2} - 64}=\frac{1}{\left(x - 8\right) \left(x + 8\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(x - 8\right) \left(x + 8\right)}=\frac{A}{x - 8}+\frac{B}{x + 8}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(x - 8\right) \left(x + 8\right)}=\frac{\left(x - 8\right) B + \left(x + 8\right) A}{\left(x - 8\right) \left(x + 8\right)}$$

The denominators are equal, so we require the equality of the numerators:

$$1=\left(x - 8\right) B + \left(x + 8\right) A$$

Expand the right-hand side:

$$1=x A + x B + 8 A - 8 B$$

Collect up the like terms:

$$1=x \left(A + B\right) + 8 A - 8 B$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + B = 0\\8 A - 8 B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{16}$$$, $$$B=- \frac{1}{16}$$$

Therefore,

$$\frac{1}{\left(x - 8\right) \left(x + 8\right)}=\frac{\frac{1}{16}}{x - 8}+\frac{- \frac{1}{16}}{x + 8}$$

Answer: $$$\frac{1}{x^{2} - 64}=\frac{\frac{1}{16}}{x - 8}+\frac{- \frac{1}{16}}{x + 8}$$$